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Ne4ueva [31]
3 years ago
9

Which expression is equivalent to (2mn)^4/6m^-3n^-2? Assume .m=0,n=0

Mathematics
2 answers:
Elza [17]3 years ago
7 0

Given expression:\frac{(2mn)^4}{6m^{-3}n^{-2}}.

\mathrm{Apply\:exponent\:rule}:\quad \left(ab\right)^c=a^cb^c

\left(2mn\right)^4:\quad 2^4m^4n^4

=\frac{2^4m^4n^4}{6m^{-3}n^{-2}}

=\frac{2^4m^4n^4}{2\cdot \:3m^{-3}n^{-2}}

=\frac{2^3m^4n^4}{3m^{-3}n^{-2}}

\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}\:=\:x^{a-b}

\frac{m^4}{m^{-3}}=m^{4-\left(-3\right)}=m^7

=\frac{2^3m^7n^4}{3n^{-2}}

\frac{n^4}{n^{-2}}=n^{4-\left(-2\right)}=n^6

=\frac{2^3m^7n^6}{3}

=\frac{8m^7n^6}{3}\

<h3>Therefore, correct option is first option \frac{8m^7n^6}{3}.</h3>

kati45 [8]3 years ago
4 0
<span>8m^7n^6
------------
     3</span>
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Math! 30 points + brainliest
Ronch [10]

Answer:

A) 9.56x10^38 ergs

B) 7.4x10^-3 mm

Step-by-step explanation:

A) 9.56x10^38 ergs B) 7.4x10^-3 mm A). For the sun, just multiply the power by time, so 3.9x10^33 erg/sec * 2.45x10^5 sec = 9.56x10^38 B) Of the two values 7.4x10^-3 and 7.4x10^3, the value 7.4x10^-3 is far more reasonable as a measurement for blood cell. Reason becomes quite evident if you take the 7.4x10^3 value and convert to a non-scientific notation value. Since the exponent is positive, shift the decimal point to the right. So 7.4x10^3 mm = 7400 mm, or in easier to understand terms, over 7 meters. That is way too large for a blood cell when you consider that you need a microscope to see one. Now the 7.4x10^-3 mm value converts to 0.0074 mm which is quite small and would a reasonable size for a blood cell.



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Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based
notsponge [240]

Answer:

The Taylor series for sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}, the first three non-zero terms are 9x^{2} -27x^{4}+\frac{162}{5}x^{6} and the interval of convergence is ( -\infty, \infty )

Step-by-step explanation:

<u>These are the steps to find the Taylor series for the function</u> sin^2(3 x)

  1. Use the trigonometric identity:

sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))

   2. The Taylor series of cos(x)

cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}

Substituting y=6x we have:

cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}

   3. Find the Taylor series for sin^2(3x)

sin^{2}(3x)=\frac{1}{2}*(1-cos(6x)) (1)

cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!} (2)

Substituting (2) in (1) we have:

\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}

Bring the factor \frac{1}{2} inside the sum

\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )

\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}

Extract the term for n=0 from the sum:

\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}

<u>To find the first three non-zero terms you need to replace n=3 into the sum</u>

sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}

<u>To find the interval on which the series converges you need to use the Ratio Test that says</u>

For the power series centered at x=a

P(x)=C_{0}+C_{1}(x-a)+C_{2}(x-a)^{2}+...+ C_{n}(x-a)^{n}+...,

suppose that \lim_{n \to \infty} |\frac{C_{n}}{C_{n+1}}| = R.. Then

  • If R=\infty, the the series converges for all x
  • If 0 then the series converges for all |x-a|
  • If R=0, the the series converges only for x=a

So we need to evaluate this limit:

\lim_{n \to \infty} |\frac{\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}}{\frac{-9^{n+1}2^{2*(n+1)-1}x^{2*(n+1)}}{(2*(2n+1))!}} |

Simplifying we have:

\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |

Next we need to evaluate the limit

\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |\\ \frac{1}{18x^{2} } \lim_{n \to \infty} |-(n+1)(2n+1)}|}

-(n+1)(2n+1) is negative when n -> ∞. Therefore |-(n+1)(2n+1)}|=2n^{2}+3n+1

You can use this infinity property \lim_{x \to \infty} (ax^{n}+...+bx+c) = \infty when a>0 and n is even. So

\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } | \\ \frac{1}{18x^{2}} \lim_{n \to \infty} 2n^{2}+3n+1=\infty

Because this limit is ∞ the radius of converge is ∞ and the interval of converge is ( -\infty, \infty ).

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